Determine the number of group homomorphisms $f:\mathbb Z_{63}\to\mathbb Z_{147}$ with $|image(f)|=7$.

Question

• Solution is apparently 6.

Current attempt, although, I am not sure whether this the right approach:

Let $f(x):=x$. We require $f(63)=63x\equiv 0 \mod 147$; that is, $63x=147k\iff 7\mid x$. So we have 21 distinct homomorphisms.

If we consider a few, say,

$f_7(a)=7a\in \mathbb Z_{147}\implies |img(f_7)|=\lfloor \frac{147}{7}\rfloor =21$,

$f_{14}(a)=14a\in \mathbb Z_{147}\implies |img(f_{14})|=\lfloor \frac{147}{14}\rfloor =10$,

$\vdots$

$f_{7t}(a)=7ta \implies |img(f_{7t})|=\lfloor \frac{21}{t}\rfloor$ for $t\in\{0,...,20\}$(verify inductively).

Thus, setting $\lfloor \frac{21}{t}\rfloor=7\implies t=3$ as the unique $t$ within our specified range. What other 5 homomorphisms with images of size 7 am I missing? Ideally, I would like a general method for finding any image size

Answer 1

There are a number of homomorphisms from $\Bbb Z_{63}$ to $\Bbb Z_{147}. $

As usual since $\Bbb Z_{63}$ is cyclic, each is determined by $h (1).$

The only requirement is that $\mid h (1)\mid\mid63.$

A cyclic group has one cyclic subgroup of every order dividing its order. There's $\varphi (7)=6$ generators for the subgroup of order $7$ in $\Bbb Z_{147}.$

Defining $h(1)$ to be any of these six gives different homomorphisms with image order $7.$

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For an image size $m$, there's $\varphi (m)$ generators to choose from, where $\varphi $ is Euler's phi function.

Any of the mutual divisors of $147$ and $63$ is a possibility. So, $m=1,3,7,$ or $21.$

The number of homomorphisms in each case is $1,2,6$ or $12.$

Answer 2

Short solution: there is only one subgroup of order $7$ in $\Bbb Z_{147}$ and since $7$ is prime, this subgroup is generated by any of its $6$ nonzero elements, so there are exactly $6$ possible choices for your $x:=f(1)$.

Solution following (and correcting) your approach: a homomorphism $f:\mathbb Z_{63}\to\mathbb Z_{147}$ is determined by the choice of $x:=f(1)\in\mathbb Z_{147}$ such that $7\mid x$. Let us denote it (like you did) by $f_{7t}$, where $x=7t$ and $t\in\{0,1,\dots,20\}$. Then, $|img(f_{7t})|$ is not equal to your $\lfloor\frac{21}{t}\rfloor$ (for instance $|img(f_{14})|$ cannot be $10$ since it must divide $147$) but to $\frac{21}{\gcd(21,t)}$, so that it equals $7$ iff $\gcd(21,t)=3$. This leaves us with $6$ solutions: $t\in\{3,6,9,12,15,18\}$.

More generally, by the same reasoning, for any of the four divisors $d=1,3,7,21$ of $21$, the number of homomorphisms $f:\mathbb Z_{63}\to\mathbb Z_{147}$ such that $|img(f)|=d$ is the number of integers $t\in\{0,1,\dots,20\}$ such that $\gcd(21,t)=\frac{21}d$, i.e. respectively $1,2,6,12$ (you may check that $1+2+6+12=21$).